SAT Math 9.5 Algebra and Functions
|Algebra II||Quadratic Equations|
|Algebra and Functions||Solving Quadratic Equations by Factoring|
|Passport to Advanced Math||Quadratic and exponential functions (word problems)|
|Product Type||SAT Math|
|Quadratic Equations||Quadratic Equations|
|SAT Math||Algebra and Functions|
Let’s take a closer look at the second requirement.
This just means that if we plugged the value (x + 1) into the function,
which we’re given in the first requirement, we can solve for x.
To solve this equation, we’re first going to have to turn it back into a polynomial.
So, we expand (x + 1) squared first.
By applying foil, we turn (x + 1) squared into x squared + 2x + 1.
Then, we distribute -6 into the second parentheses. We get -6x - 6.
Now we can combine like terms. There’s only one x squared term, so that stays by itself.
However, we have both 2x and -6x, so we can combine those to -4x.
Then, all of the constants add to +3. Great, we have a polynomial!
To be more specific… it's a quadratic.
We can just stick this puppy into the quadratic formula and come out with the answer.
Plugging in our values, we get that x is equal to 4 plus or minus the square root of 16 – 12 over 2.
This simplifies to 4 plus or minus 2 over 2. The possibilities are 1 or 3.
That's our answer either 1 or 3.